Question

数学题 用因式分解 第一题 x（x-2)=x-2 第二题 5x(x+1)=3x 第三题 （2x-3）²-4=0

第四题 x²+8=4x+2
第五题 3x(2x+1)=4x+2
第六题 (3x+4)²=7(3x+4）
本人急用
好的加分

Answer 1

x( x - 2 ) = x - 2
x( x - 2 ) - ( x - 2 ) = 0
( x - 2 )( x - 1 ) = 0
x1 = 1
x2 = 2

5x( x + 1 ) = 3x
x( 5x + 5 ) - 3x = 0
x( 5x + 5 - 3 ) = 0
x( 5x + 2 ) = 0
x1 = 0
x2 = - 2/5

( 2x - 3 )" - 4 = 0
( 2x - 3 )" - 2" = 0
( 2x - 3 - 2 )( 2x - 3 + 2 ) = 0
( 2x - 5 )( 2x - 1 ) = 0
x1 = 5/2
x2 = 1/2

x" + 8 = 4x + 2
x" - 4x + 6 = 0
x" - 4x + 4 + 2 = 0
( x - 2 )" + 2 = 0
无实数解

3x( 2x + 1 ) = 4x + 2
3x( 2x + 1 ) = 2( 2x + 1 )
3x( 2x + 1 ) - 2( 2x + 1 ) = 0
( 2x + 1 )( 3x - 2 ) = 0
x1 = -1/2
x2 = - 2/3

( 3x + 4 )" = 7( 3x + 4 )
( 3x + 4 )" - 7( 3x + 4 ) = 0
( 3x + 4 )( 3x + 4 - 7 ) = 0
( 3x + 4 )( 3x - 3 ) = 0
( x + 4/3 )( x - 1 ) = 0
x1 = 1
x2 = - 4/3

追问

可以用根号

追答

x" + 8 = 4x + 2
x" - 4x + 6 = 0
x" - 4x + 4 + 2 = 0
( x - 2 )" + 2 = 0
其实是复数
( x - 2 )" - [ (√2)i ]" = 0
[ x - 2 + (√2)i ][ x - 2 - (√2)i ] = 0
x1 = 2 + (√2)i
x2 = 2 - (√2)i